ring in which every proper ideal is primary 
Let $R$ be a commutative ring in which every proper ideal is a primary ideal. Prove that $R$ has at most one nonzero prime ideal and a proper ideal $I$ is prime if and only if $[$ for all $x \in R$ if $x^2 \in I$ then $x \in I$ $]$.

I have indicated that if the intersection of two prime ideals $P_1, P_2$ is a primary ideal then either $P_1 \subset P_2$ or $P_2 \subset P_1$. Hence, it implies that there is at most $1$ maximal ideal in $R$. My attempt is to indicate that every prime ideal in $R$ is a maximal ideal. 
 A: (NOTE I know I've seen this result somewhere before, but I can't locate it now. Please, anybody, edit with a reference if you know of one.)
I'm assuming $R$ has an identity... no idea what happens without.
To start with the second half of the question, it's always the case that a primary ideal $I$ is prime iff $x^2 \in I \implies x \in I$.  Indeed, let $ab \in I$.  By primaryness, either $a \in I$ or $b^n \in I$ for some $n$.  In the latter case, additionally $b^m \in I$ for some $m$ a power of $2$.  Repeated applications of the property yields $b^{m/2} \in I, \ldots$, etc. $b \in I$.
The other part of the question is more interesting.  
Let $R$ be a ring in which every ideal is primary.  To rehash your work so far, this entails that all radical ideals in $R$ are prime, and in turn that primeness in $R$ is preserved under intersection.  In particular, if $P$ and $Q$ are two primes, then $PQ \subseteq P \cap Q$ (which is prime) implies that $P \subseteq P \cap Q \subseteq Q$ or $Q \subseteq P \cap Q \subseteq P$.  It follows that the prime ideals of $R$ are totally ordered.  
So, let $R$ have unique maximal ideal $\mathfrak{m}$ and unique minimal ideal $\mathfrak{q}$.  We will allow that they coincide, e.g. if $R$ is a field, or, as we'll soon see, any non-domain.
Let $\mathfrak{p}$ any prime ideal and pick $x \notin \mathfrak{p}$, $a \in \mathfrak{p}$.  This gives us $axR \subseteq \mathfrak{p}$.  Since $axR$ is primary, either $a \in axR$ or $x^n \in axR$ for some $n$, and this implies $a \in axR$ (if $x^n \in axR$, then $x^n \in \mathfrak{p}$, hence $x \in \mathfrak{p}$, a contradiction.)  Thus we have $a = axc$ for some $0 \not= c \in R$, and $a(1-xc) = 0$.  This equation alone affords us the classification that we seek.
If $\mathfrak{q} \subsetneq \mathfrak{m}$, then setting $\mathfrak{p} = \mathfrak{q}$ and choosing $x \in \mathfrak{m} \setminus \mathfrak{q}$ would have $1-xc$ a unit, hence $a = 0$.  Thus $\mathfrak{q} \subsetneq \mathfrak{m}$ forces $R$ to be a domain.  On the other hand, $R$ a domain would force either $\mathfrak{p} = 0$ (if the complement of $\mathfrak{p}$ contains a nonunit) or $\mathfrak{p} = \mathfrak{m}$ (if $\mathfrak{p}$ contains a nonzero element).  
So in total, we have proved 

Let $R$ be a ring in which every ideal is primary.  Then $R$ is one of the following:
  - A ring with exactly one prime (equivalently, a ring in which every element is either a unit or nilpotent)
  -A $1$-dimensional domain with a unique maximal ideal.

Actually the converse is true too.
A ring with exactly one prime has $(0)$ a primary ideal. Quotienting by a proper ideal $I$ yields another ring with exactly one prime, hence $(0)$ is primary in $R/I$ and $I$ must have been primary.  
